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At the end of 2004, a certain farm had 24 hens, 12 cows, 30 [#permalink]
07 Mar 2012, 13:23

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Question Stats:

56% (04:16) correct
44% (02:46) wrong based on 61 sessions

At the end of 2004, a certain farm had 24 hens, 12 cows, 30 sheep, and 14 pigs. By the end of 2005, 22 new animals — each either a hen, cow, sheep or pig — were brought to the farm. No animals left the farm. How many pigs were there on the farm at the end of 2005?

(1) The ratio of cows to pigs and the ratio of hens to sheep were the same at the end of 2004 and 2005. (2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005 _________________

Re: Ratio of x-y-z-w [#permalink]
07 Mar 2012, 13:58

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This post received KUDOS

Expert's post

At the end of 2004, a certain farm had 24 hens, 12 cows, 30 sheep, and 14 pigs. By the end of 2005, 22 new animals — each either a hen, cow, sheep or pig — were brought to the farm. No animals left the farm. How many pigs were there on the farm at the end of 2005?

(1) The ratio of cows to pigs and the ratio of hens to sheep were the same at the end of 2004 and 2005.

At the end of 2004 the ratio of cows to pigs was 12/14=6/7, in order the ratio to remain the same at the end of 2005, cows and pigs should be added in the same ratio (or not added at all). So, possible numbers of new cows and pigs are: (0, 0) or (6, 7). Notice that (12, 14) is not possible since 12+14=26, which is more than total number of new animals brought, 22.

The same for the ratio of hens to sheep: at the end of 2004 the ratio was 24/30=4/5. So, possible numbers of new hens and sheep are: (0, 0), (4, 5) or (8, 10).

Only one combination makes total of 22 animals: (6, 7) and (4, 5) --> 6+7+4+5=22. Hence, 7 new pigs were brought to the farm, so there were 14+7=21 pigs on the farm at the end of 2005. Sufficient.

(2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005. 5 sheep were brought to the farm, but we know nothing about the rest of 17 animals. Not sufficient.

Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30 [#permalink]
07 Mar 2012, 14:32

Expert's post

Cool

I changed the animals with variables to make it look more difficult, muddy the waters.

I think that when you realize that you have the relationship between ratios is already clear sufficient and move on is the right thing but your deep explanation is always useful to understand.

Next time I'll post the question correctly, eventhough it was the same thing.

Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30 [#permalink]
08 Mar 2012, 04:25

it is said that we do not have to solve the DS questions. As can be seen, however, this is not the case for most of the time, especially after going through Bunuel's solutions:) Thanks for the solution.

Questtion : Total no of animals in farm currently = 80 H=24, C=1, S=30, P=14

(1) The ratio of cows to pigs and the ratio of hens to sheep were the same at the end of 2004 and 2005. C/P and H/S will be the same for 2004 and 2005 respectively.

Hence for 2004 : C/P = 12/14 = 6/7 Hence, the multiplication factor for the ratio = x =2 H/S = 24/30 =4/5 Hence, the multiplication factor for the ratio = y = 6

Since the no of animals inc by 22 and none of the animals leave the farm, the new factors say x' and y' will be such that x'>=x or y'>=y

hence, 6x'+7x'+4y'+5y' = 80+22 13x'+9y'=102

trying with the first combination say, x'=3 and y'=7 39 + 63 = 102 102 = 102 (Bingo!!)

Thus, x' = 3 and y'=7 Therefore no of pigs = 7x' = 21 so, 7 new pigs were added =>Sufficient

(2) The number of sheep increased by 1/6 from the end of 2004 to the end of 2005 We still know nothing about the other animals =>Not sufficient

Ans: A _________________

PS: Like my approach? Please Help me with some Kudos.

gmatclubot

Re: At the end of 2004, a certain farm had 24 hens, 12 cows, 30
[#permalink]
25 Jun 2013, 07:16

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...